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Viewing Single Post From: The giant space ship example
Chris Ho-Stuart
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Dec 7 2011, 06:44 AM

If you turn off the furnaces, how long would take to cool the atmosphere.
You seem to think it would happening quickly, I don't.

I haven't said anything at all about how long it would take. But I'll do so now, just quickly, for the first time.

To answer this, we need to some assumptions.

  • I'll assume atmosphere of N2 and O2. Hence no thermal emission to speak of.
  • I'll assume that the bottom of the atmosphere has a good thermal contact with the surface. Hence the bottom of the atmosphere cools as the surface cools, and the rest of the atmosphere follows.
  • I'll assume that the atmosphere stays mixed, somehow, so that we keep the adiabatic lapse rate. This is actually a bad assumption. As the bottom cools, you'll get a stable inversion. But ignore that for now.
  • I'll assume that the surface is well insulated, as it would be with a dusting of a few cm of regolith, for example. This means that we can ignore heat sinks below the surface, and just worry about the atmosphere.

From post #5 in the thread, mass of the atmosphere is 5.4e17 kg. Heat capacity Cp is 1 J/g/K, or 1000 J/kg/K. So the total atmosphere heat capacity is 5.4e21 5.4e20 J/K.

Energy being lost to space is 6.58e15 W, so the cooling rate is 1.2e-6 1.2e-5 K/s. That is, it would take 8.2e5 8.2e4 seconds to drop one degree, or about 9.5 days 23 hours. As it cools, the rate of cooling drops, as you are losing energy more slowly.

[indentblock]I've fixed up an error in my calculations, where I had the heat capacity ten times too high[/indentblock]

However, this calculation has ignored the inversion layer that would form at the bottom of the atmosphere. In practice it would drop in temperature faster than this until you get the inversion in approximate equilibrium with a slow diffusion of heat from higher altitudes. Calculating that is just more distraction from the original problem posed.


To quantify this, I think it might drop in temperature fairly quickly by a couple of degrees. And by quickly I mean in a minute or two of time.
As example in solar ellipse I have been informed [never been to one] the temperature does drop rapidly. So something like 10 degrees in couple minutes. Enough that everyone notices it. Apparently.

Yes indeed. Nighttime inversions over the desert can be very shallow, just a few meters even; and that cools much more quickly than the entire atmosphere. A solar eclipse would get a rapid inversion forming also. (I've been in a total solar eclipse, but it was heavily overcast at the time, unfortunately. It went dark like at night, but there was not much temperature drop. Nor would I expect much drop in those conditions of heavy cloud.)

Just for the sake of argument, let's assume that about the bottom 4 meters of atmospheres is the effective heat capacity as temperatures start to fall. At pressure 0.3 atm, that's about 1.2 kg of air per square meter, so the total effective heat capacity over the starship surface is roughly 2e17 J/K. Using this value in the calculation above give temperature falling initially at 1 degree every 300 seconds or so; or 5 minutes for a degree. That's a really rough ball park calculation. If you have a poor thermal contact with the surface, then ground temperature will drop much faster, but what we feel is the lowest levels of air, and that would drop more slowly.


We also have temperatures dropping by 100 K during couple hours on the Moon during a Lunar ellipse.

There's no atmosphere there; the only heat sink is the top few centimeters of regolith. I think it falls much faster than that even, but I have no figures.

So if cooled at 1 K per hour, it would require 30 times more energy
than you said.
So what you are actually saying is it will take about 30 hrs to lower 1 K- if 6,580,000 GigaWatts is correct.

This is the first post in which I have said how long cooling takes.

So what I am getting is ground surface temperature doesn't matter. Amount of atmosphere doesn't matter. In terms reducing power requirements.

The power requirements to maintain a given temperature depend crucially on the surface temperature, the amount of atmosphere, and the atmospheric composition.

The rate of cooling when you switch off the generators depends on the same things, though the initial cooling rate will also depend on thermal contact between surface and atmosphere. Any additional surface heat capacity also matters, but it would most likely be negligible.

Cheers -- Chris
Edited by Chris Ho-Stuart, Dec 7 2011, 09:02 PM.
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The giant space ship example · Physical theory for climate